Convert Decimal -0.000 000 000 000 000 603 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 603(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 000 603(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 000 603| = 0.000 000 000 000 000 603


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 603.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 000 603 × 2 = 0 + 0.000 000 000 000 001 206;
  • 2) 0.000 000 000 000 001 206 × 2 = 0 + 0.000 000 000 000 002 412;
  • 3) 0.000 000 000 000 002 412 × 2 = 0 + 0.000 000 000 000 004 824;
  • 4) 0.000 000 000 000 004 824 × 2 = 0 + 0.000 000 000 000 009 648;
  • 5) 0.000 000 000 000 009 648 × 2 = 0 + 0.000 000 000 000 019 296;
  • 6) 0.000 000 000 000 019 296 × 2 = 0 + 0.000 000 000 000 038 592;
  • 7) 0.000 000 000 000 038 592 × 2 = 0 + 0.000 000 000 000 077 184;
  • 8) 0.000 000 000 000 077 184 × 2 = 0 + 0.000 000 000 000 154 368;
  • 9) 0.000 000 000 000 154 368 × 2 = 0 + 0.000 000 000 000 308 736;
  • 10) 0.000 000 000 000 308 736 × 2 = 0 + 0.000 000 000 000 617 472;
  • 11) 0.000 000 000 000 617 472 × 2 = 0 + 0.000 000 000 001 234 944;
  • 12) 0.000 000 000 001 234 944 × 2 = 0 + 0.000 000 000 002 469 888;
  • 13) 0.000 000 000 002 469 888 × 2 = 0 + 0.000 000 000 004 939 776;
  • 14) 0.000 000 000 004 939 776 × 2 = 0 + 0.000 000 000 009 879 552;
  • 15) 0.000 000 000 009 879 552 × 2 = 0 + 0.000 000 000 019 759 104;
  • 16) 0.000 000 000 019 759 104 × 2 = 0 + 0.000 000 000 039 518 208;
  • 17) 0.000 000 000 039 518 208 × 2 = 0 + 0.000 000 000 079 036 416;
  • 18) 0.000 000 000 079 036 416 × 2 = 0 + 0.000 000 000 158 072 832;
  • 19) 0.000 000 000 158 072 832 × 2 = 0 + 0.000 000 000 316 145 664;
  • 20) 0.000 000 000 316 145 664 × 2 = 0 + 0.000 000 000 632 291 328;
  • 21) 0.000 000 000 632 291 328 × 2 = 0 + 0.000 000 001 264 582 656;
  • 22) 0.000 000 001 264 582 656 × 2 = 0 + 0.000 000 002 529 165 312;
  • 23) 0.000 000 002 529 165 312 × 2 = 0 + 0.000 000 005 058 330 624;
  • 24) 0.000 000 005 058 330 624 × 2 = 0 + 0.000 000 010 116 661 248;
  • 25) 0.000 000 010 116 661 248 × 2 = 0 + 0.000 000 020 233 322 496;
  • 26) 0.000 000 020 233 322 496 × 2 = 0 + 0.000 000 040 466 644 992;
  • 27) 0.000 000 040 466 644 992 × 2 = 0 + 0.000 000 080 933 289 984;
  • 28) 0.000 000 080 933 289 984 × 2 = 0 + 0.000 000 161 866 579 968;
  • 29) 0.000 000 161 866 579 968 × 2 = 0 + 0.000 000 323 733 159 936;
  • 30) 0.000 000 323 733 159 936 × 2 = 0 + 0.000 000 647 466 319 872;
  • 31) 0.000 000 647 466 319 872 × 2 = 0 + 0.000 001 294 932 639 744;
  • 32) 0.000 001 294 932 639 744 × 2 = 0 + 0.000 002 589 865 279 488;
  • 33) 0.000 002 589 865 279 488 × 2 = 0 + 0.000 005 179 730 558 976;
  • 34) 0.000 005 179 730 558 976 × 2 = 0 + 0.000 010 359 461 117 952;
  • 35) 0.000 010 359 461 117 952 × 2 = 0 + 0.000 020 718 922 235 904;
  • 36) 0.000 020 718 922 235 904 × 2 = 0 + 0.000 041 437 844 471 808;
  • 37) 0.000 041 437 844 471 808 × 2 = 0 + 0.000 082 875 688 943 616;
  • 38) 0.000 082 875 688 943 616 × 2 = 0 + 0.000 165 751 377 887 232;
  • 39) 0.000 165 751 377 887 232 × 2 = 0 + 0.000 331 502 755 774 464;
  • 40) 0.000 331 502 755 774 464 × 2 = 0 + 0.000 663 005 511 548 928;
  • 41) 0.000 663 005 511 548 928 × 2 = 0 + 0.001 326 011 023 097 856;
  • 42) 0.001 326 011 023 097 856 × 2 = 0 + 0.002 652 022 046 195 712;
  • 43) 0.002 652 022 046 195 712 × 2 = 0 + 0.005 304 044 092 391 424;
  • 44) 0.005 304 044 092 391 424 × 2 = 0 + 0.010 608 088 184 782 848;
  • 45) 0.010 608 088 184 782 848 × 2 = 0 + 0.021 216 176 369 565 696;
  • 46) 0.021 216 176 369 565 696 × 2 = 0 + 0.042 432 352 739 131 392;
  • 47) 0.042 432 352 739 131 392 × 2 = 0 + 0.084 864 705 478 262 784;
  • 48) 0.084 864 705 478 262 784 × 2 = 0 + 0.169 729 410 956 525 568;
  • 49) 0.169 729 410 956 525 568 × 2 = 0 + 0.339 458 821 913 051 136;
  • 50) 0.339 458 821 913 051 136 × 2 = 0 + 0.678 917 643 826 102 272;
  • 51) 0.678 917 643 826 102 272 × 2 = 1 + 0.357 835 287 652 204 544;
  • 52) 0.357 835 287 652 204 544 × 2 = 0 + 0.715 670 575 304 409 088;
  • 53) 0.715 670 575 304 409 088 × 2 = 1 + 0.431 341 150 608 818 176;
  • 54) 0.431 341 150 608 818 176 × 2 = 0 + 0.862 682 301 217 636 352;
  • 55) 0.862 682 301 217 636 352 × 2 = 1 + 0.725 364 602 435 272 704;
  • 56) 0.725 364 602 435 272 704 × 2 = 1 + 0.450 729 204 870 545 408;
  • 57) 0.450 729 204 870 545 408 × 2 = 0 + 0.901 458 409 741 090 816;
  • 58) 0.901 458 409 741 090 816 × 2 = 1 + 0.802 916 819 482 181 632;
  • 59) 0.802 916 819 482 181 632 × 2 = 1 + 0.605 833 638 964 363 264;
  • 60) 0.605 833 638 964 363 264 × 2 = 1 + 0.211 667 277 928 726 528;
  • 61) 0.211 667 277 928 726 528 × 2 = 0 + 0.423 334 555 857 453 056;
  • 62) 0.423 334 555 857 453 056 × 2 = 0 + 0.846 669 111 714 906 112;
  • 63) 0.846 669 111 714 906 112 × 2 = 1 + 0.693 338 223 429 812 224;
  • 64) 0.693 338 223 429 812 224 × 2 = 1 + 0.386 676 446 859 624 448;
  • 65) 0.386 676 446 859 624 448 × 2 = 0 + 0.773 352 893 719 248 896;
  • 66) 0.773 352 893 719 248 896 × 2 = 1 + 0.546 705 787 438 497 792;
  • 67) 0.546 705 787 438 497 792 × 2 = 1 + 0.093 411 574 876 995 584;
  • 68) 0.093 411 574 876 995 584 × 2 = 0 + 0.186 823 149 753 991 168;
  • 69) 0.186 823 149 753 991 168 × 2 = 0 + 0.373 646 299 507 982 336;
  • 70) 0.373 646 299 507 982 336 × 2 = 0 + 0.747 292 599 015 964 672;
  • 71) 0.747 292 599 015 964 672 × 2 = 1 + 0.494 585 198 031 929 344;
  • 72) 0.494 585 198 031 929 344 × 2 = 0 + 0.989 170 396 063 858 688;
  • 73) 0.989 170 396 063 858 688 × 2 = 1 + 0.978 340 792 127 717 376;
  • 74) 0.978 340 792 127 717 376 × 2 = 1 + 0.956 681 584 255 434 752;
  • 75) 0.956 681 584 255 434 752 × 2 = 1 + 0.913 363 168 510 869 504;
  • 76) 0.913 363 168 510 869 504 × 2 = 1 + 0.826 726 337 021 739 008;
  • 77) 0.826 726 337 021 739 008 × 2 = 1 + 0.653 452 674 043 478 016;
  • 78) 0.653 452 674 043 478 016 × 2 = 1 + 0.306 905 348 086 956 032;
  • 79) 0.306 905 348 086 956 032 × 2 = 0 + 0.613 810 696 173 912 064;
  • 80) 0.613 810 696 173 912 064 × 2 = 1 + 0.227 621 392 347 824 128;
  • 81) 0.227 621 392 347 824 128 × 2 = 0 + 0.455 242 784 695 648 256;
  • 82) 0.455 242 784 695 648 256 × 2 = 0 + 0.910 485 569 391 296 512;
  • 83) 0.910 485 569 391 296 512 × 2 = 1 + 0.820 971 138 782 593 024;
  • 84) 0.820 971 138 782 593 024 × 2 = 1 + 0.641 942 277 565 186 048;
  • 85) 0.641 942 277 565 186 048 × 2 = 1 + 0.283 884 555 130 372 096;
  • 86) 0.283 884 555 130 372 096 × 2 = 0 + 0.567 769 110 260 744 192;
  • 87) 0.567 769 110 260 744 192 × 2 = 1 + 0.135 538 220 521 488 384;
  • 88) 0.135 538 220 521 488 384 × 2 = 0 + 0.271 076 441 042 976 768;
  • 89) 0.271 076 441 042 976 768 × 2 = 0 + 0.542 152 882 085 953 536;
  • 90) 0.542 152 882 085 953 536 × 2 = 1 + 0.084 305 764 171 907 072;
  • 91) 0.084 305 764 171 907 072 × 2 = 0 + 0.168 611 528 343 814 144;
  • 92) 0.168 611 528 343 814 144 × 2 = 0 + 0.337 223 056 687 628 288;
  • 93) 0.337 223 056 687 628 288 × 2 = 0 + 0.674 446 113 375 256 576;
  • 94) 0.674 446 113 375 256 576 × 2 = 1 + 0.348 892 226 750 513 152;
  • 95) 0.348 892 226 750 513 152 × 2 = 0 + 0.697 784 453 501 026 304;
  • 96) 0.697 784 453 501 026 304 × 2 = 1 + 0.395 568 907 002 052 608;
  • 97) 0.395 568 907 002 052 608 × 2 = 0 + 0.791 137 814 004 105 216;
  • 98) 0.791 137 814 004 105 216 × 2 = 1 + 0.582 275 628 008 210 432;
  • 99) 0.582 275 628 008 210 432 × 2 = 1 + 0.164 551 256 016 420 864;
  • 100) 0.164 551 256 016 420 864 × 2 = 0 + 0.329 102 512 032 841 728;
  • 101) 0.329 102 512 032 841 728 × 2 = 0 + 0.658 205 024 065 683 456;
  • 102) 0.658 205 024 065 683 456 × 2 = 1 + 0.316 410 048 131 366 912;
  • 103) 0.316 410 048 131 366 912 × 2 = 0 + 0.632 820 096 262 733 824;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 000 603(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1011 0111 0011 0110 0010 1111 1101 0011 1010 0100 0101 0110 010(2)

6. Positive number before normalization:

0.000 000 000 000 000 603(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1011 0111 0011 0110 0010 1111 1101 0011 1010 0100 0101 0110 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 51 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 603(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1011 0111 0011 0110 0010 1111 1101 0011 1010 0100 0101 0110 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1011 0111 0011 0110 0010 1111 1101 0011 1010 0100 0101 0110 010(2) × 20 =

1.0101 1011 1001 1011 0001 0111 1110 1001 1101 0010 0010 1011 0010(2) × 2-51


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -51

Mantissa (not normalized):
1.0101 1011 1001 1011 0001 0111 1110 1001 1101 0010 0010 1011 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-51 + 2(11-1) - 1 =

(-51 + 1 023)(10) =

972(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 972 ÷ 2 = 486 + 0;
  • 486 ÷ 2 = 243 + 0;
  • 243 ÷ 2 = 121 + 1;
  • 121 ÷ 2 = 60 + 1;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

972(10) =

011 1100 1100(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0101 1011 1001 1011 0001 0111 1110 1001 1101 0010 0010 1011 0010 =

0101 1011 1001 1011 0001 0111 1110 1001 1101 0010 0010 1011 0010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1100 1100

Mantissa (52 bits) =
0101 1011 1001 1011 0001 0111 1110 1001 1101 0010 0010 1011 0010


Decimal number -0.000 000 000 000 000 603 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1100 1100 - 0101 1011 1001 1011 0001 0111 1110 1001 1101 0010 0010 1011 0010

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100